Subject : Structural Analysis 1

Topic : Shear Centre

Difficulty : High

$ I=I_{XX_1}-I_{XX_2}=\left[ \frac{100\times400^3}{12}\right]-\left[ \frac{(100\times20)(400-2\times20)}{12}\right]\\ I=(533.33\times10^6)-(311.04\times10^6)\\ \underline{I=222.3\times10^6mm^4}$

Distance of shear centre (e)

$e=\frac{h^2b^2t}{4I}=\frac{400^2\times100^2\times20}{4\times22.3\times10^6}=35.98mm\\ \boxed{e=35.98mm}\underline{Ans} $